Question: Divide the following complex numbers. $ \dfrac{1-7i}{-3-4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+4i}$ $ \dfrac{1-7i}{-3-4i} = \dfrac{1-7i}{-3-4i} \cdot \dfrac{{-3+4i}}{{-3+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(1-7i) \cdot (-3+4i)} {(-3-4i) \cdot (-3+4i)} = \dfrac{(1-7i) \cdot (-3+4i)} {(-3)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(1-7i) \cdot (-3+4i)} {(-3)^2 - (-4i)^2} = $ $ \dfrac{(1-7i) \cdot (-3+4i)} {9 + 16} = $ $ \dfrac{(1-7i) \cdot (-3+4i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({1-7i}) \cdot ({-3+4i})} {25} = $ $ \dfrac{{1} \cdot {(-3)} + {-7} \cdot {(-3) i} + {1} \cdot {4 i} + {-7} \cdot {4 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{-3 + 21i + 4i - 28 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{-3 + 21i + 4i + 28} {25} = \dfrac{25 + 25i} {25} = 1+i $